A ball falls from height h. After 1 second, another ball falls freely from a point 20 m below the point from where the first ball falls. Both of them reach the ground at the same time. What is the value of h

Options

A. 11.2 m
B. 21.2 m
C. 31.2 m
D. 41.2 m

Explanation

To solve the problem step by step, we will analyze the motion of both balls and set up equations based on their respective distances traveled and the time taken.

### Step 1: Define the Variables
Let:
- \( h \) = height from which the first ball is dropped.
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).
- \( t \) = time taken by the first ball to reach the ground.
- The second ball is dropped 20 m below the first ball after 1 second.

### Step 2: Equation for the First Ball
The first ball is dropped from height \( h \) with an initial velocity \( u = 0 \). The distance traveled by the first ball when it reaches the ground can be described by the equation of motion:
\[ h = ut + \frac{1}{2} g t^2 \]
Since \( u = 0 \):
\[ h = \frac{1}{2} g t^2 \tag{1} \]

### Step 3: Equation for the Second Ball
The second ball is dropped after 1 second, so it falls for \( t - 1 \) seconds. The distance it travels to reach the ground is \( h - 20 \) m. The equation for the second ball is:
\[ h - 20 = u(t - 1) + \frac{1}{2} g (t - 1)^2 \]
Again, since \( u = 0 \):
\[ h - 20 = \frac{1}{2} g (t - 1)^2 \tag{2} \]

### Step 4: Substitute and Expand
Now we will substitute \( g = 10 \, \text{m/s}^2 \) into both equations.

From equation (1):
\[ h = \frac{1}{2} \times 10 \times t^2 = 5t^2 \]

From equation (2):
\[ h - 20 = \frac{1}{2} \times 10 \times (t - 1)^2 = 5(t - 1)^2 \]
Expanding the right side:
\[ h - 20 = 5(t^2 - 2t + 1) = 5t^2 - 10t + 5 \]

### Step 5: Set Equations Equal
Now we can set the two expressions for \( h \) equal to each other:
\[ 5t^2 = 5t^2 - 10t + 25 \]
Subtract \( 5t^2 \) from both sides:
\[ 0 = -10t + 25 \]
Solving for \( t \):
\[ 10t = 25 \quad \Rightarrow \quad t = 2.5 \, \text{s} \]

### Step 6: Calculate \( h \)
Now substitute \( t \) back into equation (1) to find \( h \):
\[ h = 5t^2 = 5(2.5)^2 = 5 \times 6.25 = 31.25 \, \text{m} \]

### Final Answer
The value of \( h \) is \( 31.25 \, \text{m} \).

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